Does the action of the fundamental group of a space $X$ on the universal cover $\tilde{X}$ depend on the basepoints? Let $X$ be a topological space that has a universal cover $p:\tilde{X}\to X$. Fix a basepoint $\tilde{x}_0\in \tilde{X}$ and let $x_0=p(\tilde{x}_0)$. Then the group $G$ of deck transformations of $\tilde{X}$ is isomorphic to $\pi_1(X,x_0)$: for a loop $\gamma$, let $\tilde{\gamma}$ be the lift of $\gamma$ such that $\tilde{\gamma}(0)=\tilde{x}_0$. Then there is a unique deck transformation $\tau_{\gamma}$ such that $\tau_{\gamma}(\tilde{x}_0)=\tilde{\gamma}(1)$. The map $[\gamma]\to \tau_{\gamma}$ gives an isomorphism $\pi_1(X,x_0)\to G$. Via this isomorphism, we see that the group $\pi_1(X,x_0)$ acts on $\tilde{X}$. Does this action depend on the choice of $\tilde{x}_0 \in p^{-1}(x_0)$? For example, in the case of $\Bbb R\to S^1$, the action does not depend on the choice of a basepoint of $\Bbb R$. So I'm curious about the general case.
 A: Note that there is no canonical isomorphism between $\pi_1(X,x_0)$ and $\pi_1(X, x_1)$ for $x_0,x_1\in X$, as it depends on the homotopy class of the paths from $x_0$ to $x_1$. So we know the answer must be false, but it's worse than that.
To define the action of $\pi_1(X, x_0)$ on $\tilde{X}$, one needs to pick a lifting $\tilde{x_0}$ of $x_0$, therefore the action of $\pi_1(X, x_0)$ is not canonically defined. So the problem is ill-posed.
The reason that it all works out for $\mathbb R/S^1$ is $\pi_1(S^1)$ is abelian, so the group action doesn't really depend on the choice of the base point lifting.
To give a slightly different perspective, the Deck transformation group $G$ by defintion acts on $\tilde{X}$ canonically. For any point $\tilde{x}\in \tilde{X}$, we can define an isomorphism $\rho_{\tilde{x}}: G\rightarrow \pi_1(X,x)$ by $\rho(g) = [\text{projection of any path from } \tilde{x} \text{ to } g(\tilde{x})]$, but this depends on the choice of $\tilde{x}$, not just its projection $x$. In Hatcher's book, he carefully denoted $G(\tilde{X})\approx \pi_1(X)$ instead of " $\cong$".
However, the action is canonical and therefore natural for $\tilde{x}\in\tilde{X}$. More precisely, we can define the functor from the fundamental groupoid to the category of group actions on $\tilde{X}/X$ by sending $\tilde{x}\in \tilde{X}$ to the group action of $\pi_1(X, x)$ on $\tilde{X}/X$ through the lifting $x\rightarrow\tilde{x}$.
